Sketch the curve y=4-(x+3)^2, showing the points where the curve crosses the x-axis and any minimum or maximum points.

This equation rearranges to give -y=(x+3)^2-4, which is very similar to our curve y=(x+3)^2-4 from before. In fact, replacing y with -y in an equation is equivalent to reflecting the curve through the x-axis. We then take the points (-5,0), (-1,0) and (-3,-4) from before and replace y with -y, giving (-5,0), (-1,0) and (-3,4). We have found where the new curve crosses the x-axis and its minimum/maximum. The graph is an inverted u-shape since we have a -x^2 in the equation so (-3,4) is a maximum point.

JI

Related Maths GCSE answers

All answers ▸

what is the median, mode and mean?


Solve the simultaneous equations: 3y + 2x = 10, 4x - y + 3 = 2


How do you expand brackets?


If 4x = 9y, and 2x + 5y = 95. What is the value of y^2 - x?